Conic Sections (Parabola and Ellipse) - P5

More P5 work (it just keeps on coming... think I'm starting to understand bits of it, though). Here are a few questions on coordinate systems (AKA conic sections) on the parabola and the ellipse that I couldn't work out how to do:

1) Show that the line with equation 4y - 3x = 20 is a tangent to both the circle with equation x² + y² = 16 and parabola with equation y² = 15x.

I really didn't know where to start this one (I suppose you rearrange the tangent equation to eliminate y or x in another one, but it didn't work when I tried it...)

2) The eccentric angle corresponding to the point (2,1) on the ellipse with equation x² + 9y² = 13 is θ. Find tanθ. (ANS: tanθ=3/2)

I think I understood all of the theory for this question, but I got the answer tanθ = ½ :

x = acosθ and y = bsinθ
2 = acosθ 1 = bsinθ at the point (2,1) on the ellipse.

so 2bsinθ = acosθ
4b²sin² θ = a²cos²θ
4a²(1-e²)sin²θ = a²cos²θ
4sin²θ = cos²θ (as a ≠ 0)
4sin²θ = 1 - sin²θ

so sin²θ = 1/5
tanθ = 1/2

Is the above totally flawed (in the theory) or is there a specific mistake or is the answer given in the back of the book wrong? One of these has got to be the case!

3) An ellipse has focus S (√5,0) and equation x²/9 + y²/4 = 1
The variable point T(3cost, 2sint) is joined to S. The line ST is produced to P so that ST/SP = 1/3. Find the locus of P as t varies.

For this, I got confused very early - I don't really know how to start (i.e. how to find the line ST)... Is it just x = 3cost - √5 and y = 2sint ??

4) Show that the line y = 2x+5 is a tangent to the ellipse with equation 9x² + 4y² = 36. Find an equation of the corresponding normal to the ellipse.

Perhaps drawing this would help (it's something that I don't usually bother to do - is it recommendable in this sort of question??) Again, it's the idea of working backwards when you have a tangent to proving that it's from a certain ellipse that I don't understand, although I can do it the other way around...

5) The point P (7cost, 5sint) is on the ellipse with equation x²/49 +y²/25 = 1. The line through P parallel to the y-axis meets the x-axis at X. The point Q is on the line XP produced so that XQ = 2XP. Find, in cartesian form, an equation of the locus of Q as t varies.

Ok, here's my working for this (I'm fairly sure I know where I've gone wrong on this one, but I don't know how to put it right...)

If the line is parallel to the y-axis, there will be no change in y-coordinate so x = 7cost is the line (this can't be right, as the form for y isn't linear so it will change as t changes. Eurgh.)

so t = pi/2. Then I gave up becausE I was certain (still am) that this first line isn't correct.

If anyone could help me, I'd be super grateful (these boards really are a great resource for failed further maths students like me). And, of course, for talented real mathmaticians and a whole lot of other people besides.

Woo.

Reply 1

m:)ckel

Yttrium

1) Show that the line with equation 4y - 3x = 20 is a tangent to both the circle with equation x² + y² = 16 and parabola with equation y² = 15x.

Sub the equation of the tangent into each equation, and you should get just 1 solution for each (since it only touches the curve)

y = (3x+20)/4

x² + [(3x+20)/4]² = 16
x² + (9x²+120x+400)/16 = 16
25x² + 120x + 144 = 0
(5x+12)² = 0

=> 1 solution

And do the same for the other one...

Reply 2

Yttrium

Thanks for that, Mockel. :smile:

Any ideas with the other questions, anyone?

Reply 3

m:)ckel

5) P(7cost , 5sint)
They say X is a point vertically below P, lying on the x-axis. Hence, X(7cost , 0)
They also say, XQ = 2XP, where Q is a point produced on the line XP. Well, that just means XP = PQ. So, Q(7cost , 10sint)
Now consider the point Q.

x = 7cost
x/7 = cost

y = 10sint
y/10 = sint

cos²t + sin²t = 1

=> x²/49 + y²/100 = 1

Reply 4

dvs

2.
1/asint = b/a (QN/PN = b/a)
sint = 1/b
sin²t = 1/b² = 9/13 (from the equation of the ellipse)
cos²t = 4/13
tan²t = sin²t/cos²t = 9/4
tant = 3/2, as required.

3.
ST² = (3cost - sqrt[5])² + (2sint)²
This is the magnitude of the vector:
(3cost - sqrt[5])i + (2sint)j

So SP (in vector form) can be found:
SP = 3ST = (9cost - 3sqrt[5])i + (6sint)j

Draw a vector triangle and you can see that:
SP = SO + OP
OP = SP - SO = SP + OS
= (9cost - 3sqrt[5])i + (6sint)j + (sqrt[5])i
= (9cost - 2sqrt[5])i + (6sint)j

Now:
x = 9cost - 2sqrt[5] => cost = (x+2sqrt[5])/9
y = 6sint => sint = y/6

Since sin²t+cos²t=1, we have:
(x+2sqrt[5])/81 + y²/36 = 1

4.
Plug y into the equation of the ellipse and show that b²-4ac=0.
The gradient of the normal is -1/(gradient of the tangent).